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<big><big>Hello election methods members, Vote Fair,<br>
<br>
A while ago, on request by Kristofer, I introduced a method of
Binomial STV, which, unless extensive testing prove otherwise,
should be equally applicable to single as well as multiple seat
elections, being a more general theory of choice, generalising
on traditional STV and Meek method. It doesnt depend on
Condorcet pairing, a points system (descendants of Borda
method), or premature exclusion (IRV and conventional STV).<br>
<br>
Yours sincerely,<br>
Richard Lung.<br>
</big></big><br>
<br>
On 15/10/2015 23:52, VoteFair wrote:
<blockquote cite="mid:56202E10.3040203@VoteFair.org" type="cite">On
10/14/2015 10:41 AM, steve bosworth wrote: <br>
> [...] below, Steve Eppley has just now relevantly replied to
my <br>
> questions about MAM and VoteFair. As a result, do you see any
reason not <br>
> to prefer MAM over VoteFair popularity ranking given Eppley's
<br>
> explanation of how VoteFair (Kemeny-Young) is both vulnerable
to clone <br>
> attacks and requires a ‘combinatorial explosion of possible
orders of <br>
> finish’ when used to elect one of many candidates? <br>
<br>
For the purposes of electing a US president, the Condorcet-Kemeny
method, the Maximize Affirmed Majorities (MAM) method, the
Condorcet-Schulze method, the Ranked Pairs method, and any of the
pairwise-counting (especially Condorcet) methods, would ALWAYS
produce the same result! <br>
<br>
The situations in which these methods can produce different
results involve a much smaller number of voters. <br>
<br>
When a large number of voters are involved, or when a very large
number of voters are involved (as in US presidential elections),
the conditions that can create differences between these methods
do not occur. <br>
<br>
Differences between these methods can arise when ballots have
circular ambiguity. This circular preference is analogous to the
concept in the game "rock paper scissors." In politics this means
that about one third of the voters would mark a 1-2-3 ballot with
something like Trump, Bush, Clinton, and another third of the
voters would mark their ballot Bush, Clinton, Trump, and another
third of the voters would mark their ballot Clinton, Trump, Bush.
Please read these sequences carefully and notice that other
specific ballot preferences -- namely Trump, Clinton, Bush, and
Bush, Trump, Clinton, and Clinton, Bush, Trump -- are missing.
Such preferences do not occur in political situations that involve
a large number of voters. Yet this is the unusual pattern that is
presented in the example below where the Condorcet-Kemeny method
is being criticized. <br>
<br>
As a further perspective about the number of ballots involved, in
the American idol polls that I host at VoteFair.org, circular
ambiguity sometimes shows up after a few people have voted, but
circular ambiguity always disappears when the number of ballots
increases. <br>
<br>
Of course from a mathematical perspective, if a majority of US
voters intentionally want to create a situation that has circular
ambiguity -- that is, none of the candidates is a Condorcet winner
(which means that none of them is pairwise preferred over all the
other candidates) -- then this situation can be achieved, but only
if there is a high degree of cooperation, such as marking ballots
based on a central recommendation that, in turn, is based on
something like the first letter of their last name. <br>
<br>
To put this another way, keep in mind that vulnerability to clones
only affects the Condorcet-Kemeny method in rare circumstances.
Emphasis on the word rare, especially for political races. <br>
<br>
The MAM method, the Condorcet-Schulze method, the Ranked Pairs
method, and other pairwise-counting methods also have
vulnerabilities. But so far no one has estimated the percentage
of time in which these failures occur. Expressed another way,
current comparisons focus on pass-fail criteria, without measuring
how often, or how rare, those failures occur. And new criteria
are being added as voting becomes better studied. <br>
<br>
To keep this discussion further in perspective, the gap between
pairwise-counting methods and instant-runoff voting is huge. This
means that if instant-runoff voting were used in a US presidential
election, the results could easily be quite different from the
results of using any of the pairwise-counting methods. <br>
<br>
The fact that you want to use something like instant-runoff voting
in your APR method means that you are wise to recognize that a
single-winner election should use pairwise counting. The subtle
differences between pairwise-counting methods are far less
important within this perspective. <br>
<br>
Regarding the computation time for the Condorcet-Kemeny method, if
there were 130 candidates in the same race (as happened when
Arnold Schwarzenegger became the governor of California), a
cursory look at the numbers would reveal which candidates have a
possibility of winning, and that would reduce the race to no more
than ten candidates, and the computations for ten candidates can
be done within a few minutes using just one computer. And that's
using the slow approach of calculating a ranking score for every
possible sequence of those top candidates. When the calculations
are done efficiently, the elapsed time is just a few seconds. <br>
<br>
As long as you don't promote approval voting or instant-runoff
voting for use in US presidential elections I'm not going to argue
about which kind of pairwise-counting method should be used in US
presidential elections. (Here I'm talking about general
elections. Approval voting would be acceptable for primary
elections.) <br>
<br>
My bigger concern is using better methods for electing legislators
and parliament members and Congressmen. For that I've developed
other aspects of VoteFair ranking that go way beyond single-winner
methods. <br>
<br>
Again, I hope this feedback is helpful. <br>
<br>
(I think I've caught up with answering your questions. If not,
please repeat the question.) <br>
<br>
Richard Fobes <br>
<br>
<br>
<br>
On 10/14/2015 10:41 AM, steve bosworth wrote: <br>
<blockquote type="cite">Dear Richard, <br>
<br>
I know you are trying to find the time to reply to my earlier
email. <br>
However, below, Steve Eppley has just now relevantly replied to
my <br>
questions about MAM and VoteFair. As a result, do you see any
reason not <br>
to prefer MAM over VoteFair popularity ranking given Eppley's <br>
explanation of how VoteFair (Kemeny-Young) is both vulnerable to
clone <br>
attacks and requires a ‘combinatorial explosion of possible
orders of <br>
finish’ when used to elect one of many candidates? <br>
Regards, <br>
Steve Bosworth <br>
Dear Steve Eppley, <br>
<br>
Thank you for being so helpful in explaining both exactly how
MAM works <br>
and how it is superior to Kemeny-Young, i.e. KY’s vulnerability
to clone <br>
attacks and its requiring a ‘combinatorial explosion of possible
orders <br>
of finish’ when used to elect one of many candidates.Thank you
for <br>
giving priority to clarity over brevity.In this spirit, you may
also <br>
wish to consider some of the explanatory additions I have made
[within <br>
square brackets] in your text below.At least, these helped me to
be <br>
surer that I was understanding you correctly.As a result, for
example, I <br>
am now convinced that MAM should be used to elect the US
President.I <br>
know that this would require a US constitutional amendment. <br>
<br>
<br>
On a different but related question, I would very much
appreciate <br>
receiving any criticisms or suggestions you might have regarding
the <br>
very different electoral method I propose for electing the 435 <br>
congresspersons, a change that would also require a suitable US
<br>
constitutional amendment. I call this ‘new’ electoral system <br>
Associational Proportional Representation (APR) and its details
are <br>
explained in the attachment. <br>
<br>
<br>
Briefly stated, APR would allow each citizen to guarantee that
their one <br>
vote will continue mathematically to count in the House of <br>
Representatives through the ‘weighted vote’ given to the
congressperson <br>
they had helped to elect.Each weighted vote would be exactly
equally to <br>
the number of citizens in the whole country who had helped to
elect each <br>
rep.A modified single transferable voting (STV) method would
firstly be <br>
used during APR’s primary election to discover all the
‘electoral <br>
associations’ (geographically defined or not) who would be
allowed to <br>
send congresspersons to the House.APR’s later general election
would use <br>
a similar STV method to discover how many ‘weighted votes’ each
<br>
‘association’s’ elected candidate(s) would have in the House.APR
also <br>
makes a limited use of ‘Asset’ voting to guarantee that no
citizen’s <br>
vote need be wasted. <br>
<br>
<br>
I am aware that because STV reduces to IRV, it is theoretically
<br>
vulnerable to strategic voting.However, I do not yet see how
anyone <br>
could acquire the necessary for knowledge of how other people in
the <br>
whole country will vote using APR to enable any strategizer to
be <br>
successful. Consequently, I assume that any such criticism of
APR would <br>
only an attempt to apply what you call ‘aesthetic consistency
criteria’. <br>
<br>
<br>
I look forward to your comments. <br>
<br>
<br>
Best regards, <br>
<br>
<br>
Steve <br>
<br>
<br>
Stephen Bosworth <br>
<br>
<br>
------------------------------------------------------------------------
<br>
<br>
Subject: Re: MAM: A Question from Steve Bosworth <br>
To: <a class="moz-txt-link-abbreviated"
href="mailto:stevebosworth@hotmail.com">stevebosworth@hotmail.com</a>
<br>
From: <a class="moz-txt-link-abbreviated"
href="mailto:SEppley@alumni.caltech.edu">SEppley@alumni.caltech.edu</a>
<br>
Date: Sun, 4 Oct 2015 22:25:23 -0400 <br>
<br>
Hi Steve Bosworth, <br>
<br>
MAM isn't the same as Kemeny-Young unless there are fewer than 4
<br>
candidates. (Many pairwise voting methods behave the same with
fewer <br>
than 4 candidates; another method that behaves the same with
fewer than <br>
4 is Schulze's method.) <br>
<br>
A criterion that MAM satisfies but KY doesn't is Independence of
Clone <br>
Alternatives. (If you're unfamiliar with Independence of Clone <br>
Alternatives, the pair of examples below will help to explain
it.) A <br>
criterion that KY satisfies but MAM doesn't is Weak
Reinforcement: If <br>
two collections of votes tallied separately produce the same
order of <br>
finish, then that same order of finish must be produced when the
votes <br>
are tallied together.[SB:Not important for electing a
president.] <br>
<br>
Independence of Clone Alternatives is important because it's
easy for a <br>
small minority to nominate clones, whereas Weak Reinforcement is
<br>
unimportant because a simple rule can prevent a minority from
dividing <br>
the voters into separate groups (districts). Weak Reinforcement
is an <br>
example of criteria I call "aesthetic consistency" criteria;
society <br>
won't be harmed by voting methods that fail those criteria. Too
much of <br>
the academic literature of social choice theory is about
unimportant <br>
aesthetic consistency criteria. <br>
<br>
To illustrate how Kemeny-Young and MAM work, here's a pair of
examples <br>
that together also serve to illustrate the Independence of Clone
<br>
Alternatives criterion: <br>
<br>
Example 1. Suppose there are 100 voters and 3 candidates A, B
and C. <br>
Suppose the voters' top-to-bottom rankings are: <br>
<br>
40 35 25 <br>
A B C <br>
B C A <br>
C A B <br>
<br>
Kemeny-Young defines the "best" order of finish as the order of
finish <br>
that minimizes the sum of voters' pairwise preferences that the
order <br>
"disagrees" with. <br>
<br>
There are three majorities: <br>
75 rank B over C (with 25 opposed). <br>
65 rank A over B (with 35 opposed). <br>
60 rank C over A (with 40 opposed). <br>
<br>
KY says the best order of finish is ABC. ABC disagrees with the
25 who <br>
rank C over B and the 35 who rank B over A and the 60 who rank C
over A. <br>
Thus the sum of pairwise preferences that ABC disagrees with is
120. <br>
(All other possible orders of finish have a larger sum, and are
<br>
therefore worse, according to KY.) Since ABC is the order of
finish, the <br>
winner is A. <br>
<br>
<br>
<br>
Example 2: Same scenario as example 1, but suppose a clever
voter who <br>
favors B decides to also nominate a fourth alternative D that's
similar <br>
to C (but slightly inferior to C). Now the voters' top-to-bottom
<br>
rankings are: <br>
<br>
40 35 25 <br>
A B C <br>
B C D <br>
C D A <br>
D A B <br>
<br>
(Actually, let's not assume that every voter ranks C over D.
Placing D <br>
below C in all the votes above made it easier for me to write
the votes.) <br>
<br>
Now there are six majorities: <br>
C is ranked over D by a majority of size x. (x doesn't matter
but let's <br>
suppose it's largest.) <br>
75 rank B over C. <br>
75 rank B over D. <br>
65 rank A over B. <br>
60 rank Cover A. <br>
60 rank D over A. <br>
<br>
Now KY says the best order of finish is BCDA. BCDA disagrees
with] D <br>
over C and the 25 who rank C over B and the 25 who rank D over B
and the <br>
65 who rank A over B and the 40 who rank A over C and the 40 who
rank A <br>
over D, which sum to 295-x [i.e. 295 – 100]. ABCD is not the
best order <br>
of finish because it disagrees with [35 who rank B over A]the
100-x who <br>
rank D over C and[ [the 25 who rank C over B] and the 25 who
rank D over <br>
B and the [60 who rank C over A] 35 who rank B over A and the 60
who <br>
rank D over A 60 who rank C over A, which sum to 305-x. The
larger sum <br>
makes ABCD worse than BCDA, according to KY. <br>
<br>
Now the KY winner is B. The clever voter who favors B changed
the winner <br>
by nominating D. When small groups vote, typically the rules
allow one <br>
or two people to nominate an alternative, which means it's easy
for a <br>
tiny minority to manipulate the outcome when a voting method
isn't <br>
independent of clones. <br>
<br>
In example 1, MAM agrees that ABC is the best order of finish.
In <br>
example 2, MAM says the best order of finish is ABCD. (With MAM
the <br>
clever voter fails to change the winner.) Here's how MAM works:
<br>
<br>
MAM constructs the order of finish a piece at a time, by
considering <br>
the majorities one at a time, from largest majority to smallest
majority: <br>
For each majority, MAM places their more-preferred candidate
ahead <br>
of their less-preferred candidate in the order of finish, unless
MAM <br>
has already placed their less-preferred candidate ahead of their
<br>
more-preferred candidate (due to the "transitivity" property of
an <br>
order of finish, as the following will explain). <br>
<br>
In example 1, the largest majority rank B over C, so MAM places
B ahead <br>
of C in the order of finish: <br>
BC <br>
The second largest majority is A over B, so MAM next places A
ahead of B <br>
in the order of finish: <br>
ABC <br>
Because A finishes ahead of B and B finishes ahead of C, it
follows that <br>
A finishes ahead of C too. (This is the transitivity property I
mentioned.) <br>
The third largest majority is C over A, but MAM does not place C
ahead <br>
of A because A is already [transitively] ahead of C.[A>B>C
is more <br>
popular this sequence is only opposed by 60 preferences while
C>A>B is <br>
opposed by 75 + 65 = 140 preferences.At the same time, this 140
is the <br>
‘Maximal Affirmed Majority’ for the finishing sequence of
A>B>C] <br>
The completed order of finish is ABC, and MAM elects A. <br>
<br>
In example 2, let's assume the largest majority is the voters
who rank C <br>
over D (although this doesn't matter). Thus this is the first
majority <br>
that MAM considers, and MAM places C ahead of D in the order of
finish: <br>
CD <br>
<br>
Next MAM considers the second largest majority, which is either
the 75 <br>
who rank B over C or the 75 who rank B over D. Later I'll
discuss how <br>
MAM chooses between majorities that are the same size; for now
let's <br>
assume MAM decides the second largest majority is the 75 who
rank B over <br>
D and the third largest majority is the 75 who rank B over C. <br>
<br>
Since the second largest majority rank B over D, MAM now places
B ahead <br>
of D in the order of finish: <br>
CD <br>
BD <br>
<br>
Since the third largest majority rank B over C, MAM next places
B ahead <br>
of C in the order of finish: <br>
BCD <br>
[Again, for simplicity, I have picked the follow majority as the
4^th <br>
one even though it is also tied with the 65 who ranked D>A,
the 65 who <br>
ranked B>C, and the 65 who ranked B>D.] <br>
<br>
<br>
The fourth largest majority [I have similarly chosen for
simplicity] is <br>
the 65 who rank A over B, so MAM places A ahead of B in the
order of finish: <br>
ABCD <br>
Note that because A is ahead of B and B is ahead of C & D,
this means A <br>
is ahead of C & D too. (The transitivity property.) <br>
<br>
(The order of finish is now a linear ordering, ABCD, so if we
wished we <br>
could stop here and elect A. But let's keep going.) <br>
<br>
The fifth largest majority is either the 60 who rank C over A or
the 60 <br>
who rank D over A. Again I'll defer an explanation of this
"tiebreaking" <br>
of same-size majorities; for now let's assume MAM considers the
fifth <br>
largest majority to be the 60 who rank C over A and the sixth
largest <br>
majority to be the 60 who rank D over A. <br>
<br>
The fifth largest majority rank C over A, but since A is already
ahead <br>
of C in the order of finish, MAM does not place C ahead of A. <br>
<br>
The sixth largest majority rank D over A, but since A is already
ahead <br>
of D in the order of finish, MAM does not place D ahead of A. <br>
<br>
The completed order of finish is ABCD, so MAM elects A.
Nominations of <br>
similar alternatives (D, which is similar to C) don't change the
outcome <br>
with MAM. <br>
<br>
There are other criteria satisfied by MAM but not by KY, but as
far as I <br>
know the only criterion satisfied by KY but not by MAM is the
Weak <br>
Reinforcement criterion I mentioned above. When Peyton Young
promoted KY <br>
in his papers and at least one book (Equity in Theory and
Practice), he <br>
touted a criterion he called Local Independence of Irrelevant <br>
Alternatives... which is satisfied by MAM. <br>
<br>
(There's another criterion called Reinforcement by Donald Saari,
which I <br>
call Strong Reinforcement, that neither KY nor MAM satisfy: If
two <br>
collections of votes separately elect X, then X must be elected
if the <br>
votes are combined. The Borda method satisfies it, and so does
Plurality <br>
Rule. Like Weak Reinforcement, I believe Strong Reinforcement is
<br>
unimportant, because it's simple to have a rule that prevents a
minority <br>
from dividing the voters into groups.) <br>
<br>
Another problem with KY is that it takes a long time to find its
best <br>
order of finish when there are many candidates. There's a
"combinatorial <br>
explosion" of possible orders of finish, and each one needs to
be <br>
checked. The difficulty in checking all possible orders of
finish, to <br>
calculate each one's sum, is why I didn't include KY in software
I wrote <br>
years ago that compares voting methods head-to-head to see which
<br>
methods' winners are preferred by more voters. (My software
generates a <br>
series of random votes, tallies the votes using two voting
methods and <br>
accumulates statistics. I presume James Green-Armytage used
similar <br>
software to conduct his analysis.) <br>
<br>
* * * <br>
<br>
Regarding tiebreaking in MAM: <br>
<br>
There are two places in MAM's algorithm where tiebreaking can be
<br>
required. One is when majorities are the same size, as mentioned
above. <br>
The other can result from a tied pairing; the simplest example
is a <br>
two-candidate election in which the voters split 50/50. In this
case, <br>
consideration of all the majorities may not suffice to place one
<br>
candidate ahead of the other. (Transitivity may do it [as it
might be as <br>
illustrated by the following example:]. For example: suppose the
"A <br>
versus B" pairing is tied, and [at the same time there is] a
majority <br>
rank A over C, and a majority rank C over B. [These additional
pairings <br>
produce: A>C>B.] The order of finish after MAM considers
the[se] two <br>
majorities is ACB, and no tiebreaking is needed to resolve A
versus B.) <br>
<br>
In an election with many voters, it will be very rare that two
[pairing] <br>
majorities are the same size, and [thus] it will be very rare
that a <br>
pairing is tied. So the simple definition of MAM provided above
would <br>
typically suffice for public elections. <br>
<br>
The tiebreak procedure has been carefully chosen so that MAM
completely <br>
satisfies criteria such as Independence of Clone Alternatives,
Strong <br>
Pareto, etc. In elections with many voters, tiebreaking would be
<br>
extremely rare so the [complete satisfaction of these criteria
would not <br>
be in doubt’ completeness of the criteria satisfaction would not
matter, <br>
but in voting by small groups it's important to [provide a
tiebreaking <br>
method that would] completely satisfy Independence of Clones.
With any <br>
voting method that doesn't completely satisfy Independence of
Clones, <br>
elections in small groups can degenerate into farces, because
each <br>
faction will always have an incentive to nominate a huge number
of <br>
clones because there's a chance it can help and it can't
hurt.[SB:It can <br>
help by effectively multiplying the size of the ‘majority’ of
the most <br>
favored clone over the other clones.??????The more supports of
the <br>
favored clone explicitly favor her over the all of her slightly
inferior <br>
clones, the greater will be her majority.This enables each
strategizer’s <br>
vote to count (have more weight) than each other voter.] <br>
<br>
The tiebreak procedure begins the same way for both same-size
majorities <br>
and resolving tied pairings. A tiebreak ordering of the
candidates is <br>
constructed by randomly picking [one ballot]at a time, and
including <br>
into the tiebreak ordering all of [that] ballot's preferences
that don't <br>
conflict with the preferences already included from the ballots
[that <br>
may have been] previously picked [in order to break any earlier
<br>
ties.This procedure is repeated] until the tiebreak ordering is
<br>
complete. Here's a simple example: Suppose there are 3 voters
and 4 <br>
candidates A,B,C,D, and suppose the voters' top-to-bottom
rankings are <br>
as follows: <br>
<br>
A B C <br>
B=C C D <br>
D A <br>
B <br>
<br>
The middle vote [Voter 2 has] omitted A & D. MAM treats it
the same as <br>
if the voter had [explicitly] ranked A & D equally at the
bottom: <br>
<br>
A B C <br>
B=C C D <br>
D A=D A <br>
B <br>
<br>
MAM picks one of the votes at random. Suppose it's the middle
vote. MAM <br>
includes its preferences into the [1^st suggested] tiebreak
ordering, <br>
which becomes BC(A=D), identical to the [ordering of this picked
voter]. <br>
The tiebreak ordering isn't yet complete because it's not a
linear <br>
ordering (it ranks A & D the same) so MAM picks another vote
at random. <br>
Suppose it's the left vote. The left vote ranks A over D, so MAM
<br>
includes the A over D preference into the tiebreak ordering,
which [now] <br>
becomes BCAD. Now the tiebreak ordering is a linear ordering, so
there's <br>
no need to pick any more votes [at random]. <br>
<br>
Let's return to example 2, which has two pairs of same-size
majorities: <br>
two majorities are 75 voters, and two majorities are 60 voters.
When MAM <br>
reaches the point at which it wants to consider the second
largest <br>
majority, it sees that two majorities have 75 voters. So it
checks the <br>
sizes of their opposing minorities. If one of the same-size
majorities <br>
is opposed by a smaller minority, then that would be the
majority that <br>
MAM considers next. In example 2, though, the opposing
minorities are <br>
both the same size (25). So MAM picks a vote at random to start
<br>
constructing the tiebreak ordering. Suppose the vote that MAM
picks is <br>
one of the 35 votes that rank B over C over D over A. All of
this vote's <br>
preferences are included into the tiebreak ordering, which
becomes BCDA <br>
(identical to the vote that was picked). The tiebreak ordering
is now a <br>
linear ordering, so no more votes need to be picked. Next MAM
compares <br>
the two same-size majorities by checking the positions of their
<br>
less-preferred candidates in the tiebreak ordering: The
less-preferred <br>
candidate of the 75 who rank B over C is C, and the
less-preferred <br>
candidate of the 75 who rank B over D is D. Since D is behind C
in the <br>
BCDA tiebreak ordering, MAM will consider [and give priority to]
the 75 <br>
who rank B over D before considering the 75 who rank B over C.
In other <br>
words, the second largest majority is the 75 who rank B over D,
and the <br>
third largest majority is the 75 who rank B over C. <br>
<br>
Later, when MAM reaches the point where it wants to consider the
fifth <br>
largest majority, it finds 60 who rank C over A and 60 who rank
D over <br>
A. In both of these same-size majorities, the less-preferred
candidate <br>
is A. So MAM checks the positions of their more-preferred
candidates in <br>
the tiebreak ordering. The more-preferred candidate of the 60
who rank C <br>
over A is C, and the more-preferred candidate of the 60 who rank
D over <br>
A is D. Since C is ahead of D in the BCDA tiebreak ordering, MAM
will <br>
consider [and give priority to] the 60 who rank C over A before
<br>
considering the 60 who rank D over A. In other words, the fifth
largest <br>
majority is the 60 who rank C over A, and the sixth largest
majority is <br>
the 60 who rank D over A. <br>
<br>
In the case where a pairing is tied and not resolved by
transitive <br>
majorities, MAM uses the [??? following]and give priority
tiebreak <br>
ordering to break the tie(s) in the order of finish. Suppose
there's a <br>
5-candidate election and that after all the majorities have been
<br>
considered, the order of finish is (A=B)C(D=E). In other words,
A & B <br>
are tied for first place, and D & E are tied for last place.
MAM will <br>
pick votes at random to construct a linear tiebreak ordering as
<br>
described above. Suppose the tiebreak ordering is CEBDA. MAM
breaks the <br>
ties beginning with the most significant tie, in this case A
versus B. <br>
Since B is ahead of A in the tiebreak ordering CEBDA, MAM places
B ahead <br>
of A, and the order of finish becomes BAC(D=E). The next tie to
break is <br>
D versus E. Since E is ahead of D in the tiebreak ordering
CEBDA, MAM <br>
places E ahead of D, and the order of finish becomes BACED. MAM
always <br>
constructs a linear order of finish, resolving all ties. <br>
<br>
Because voters can express indifference between candidates, it's
<br>
theoretically possible that every vote will be indifferent
between two <br>
candidates. (Identical twins?) In other words, the procedure of
randomly <br>
picking votes may fail to produce a linear tiebreak ordering
even after <br>
every vote has been picked. In this case, the incomplete
tiebreak <br>
ordering is completed (made linear) by randomly completing it.
Then any <br>
ties can be broken. <br>
<br>
There's a shortcut in the tiebreak procedure that can save labor
in some <br>
cases, but it's not worth discussing the shortcut here. (Here's
a hint: <br>
the tiebreaking ordering only needs to be as linear as
necessary.) <br>
<br>
Did I answer all your questions? <br>
<br>
Would it help if I provide source code for MAM written in the
Ruby <br>
programming language? I tried to write it in a way that will
make it as <br>
easy as possible for people unfamiliar with Ruby to understand
it. It's <br>
not a complete program; it doesn't have the routines needed to
input <br>
votes or parse the votes, but those routines aren't unique to
MAM; <br>
they're needed in any pairwise voting method. <br>
<br>
Also, you can freely use the online MAM server at: <a
class="moz-txt-link-freetext" href="http://MAM.hostei.com">http://MAM.hostei.com</a>
<br>
<a class="moz-txt-link-rfc2396E" href="http://mam.hostei.com/"><http://mam.hostei.com/></a>
<br>
It has a data entry box into which you can paste all the votes.
Clicking <br>
a button underneath the data entry box causes the server to
tally the <br>
votes and construct the order of finish. You can paste all the
votes in <br>
a single copy/paste operation. Here's example 2 in a format
ready to be <br>
copy/pasted: <br>
<br>
40: A B C D <br>
35: B C D A <br>
25: C D A B <br>
<br>
There are a few interesting examples at the bottom of the
webpage, that <br>
can be copied/pasted. And of course you can invent your own. Or
use real <br>
votes from real groups of people. <br>
<br>
There's at least one other website that supposedly provides the
ability <br>
to tally votes using MAM, but I haven't seen their software and
they <br>
never asked to consult with me on the details. Mike Ossipoff
once showed <br>
me an order of finish that had a tie in it, which means it could
not be <br>
a correct implementation of MAM. <br>
<br>
Best wishes, <br>
Steve Eppley <br>
</blockquote>
<br>
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</blockquote>
<br>
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